When the current in a coil changes from 2A to 4A in 0.05 sec, the e.m.f. developed in the coil is 8V. The coefficent of self induction of the coil is

Options

A. 0.1 H
B. 0.2 H
C. 0.4 H
D. 0.3 H

Explanation

To find the coefficient of self-induction (L) of the coil, we can use the formula for induced electromotive force (e.m.f.) in a coil, which is given by:

\[ E = L \frac{Di}{Dt} \]

Where:
- \( E \) is the induced e.m.f. (in volts)
- \( L \) is the self-inductance (in henries)
- \( Di \) is the change in current (in amperes)
- \( Dt \) is the change in time (in seconds)

### Step-by-Step Solution:

1. **Identify the given values**:
- Initial current, \( I_1 = 2A \)
- Final current, \( I_2 = 4A \)
- Change in current, \( Di = I_2 - I_1 = 4A - 2A = 2A \)
- Change in time, \( Dt = 0.05s \)
- Induced e.m.f., \( E = 8V \)

2. **Substitute the values into the formula**:
\[ E = L \frac{Di}{Dt} \]
Plugging in the values we have:
\[ 8 = L \frac{2}{0.05} \]

3. **Calculate \( \frac{Di}{Dt} \)**:
\[ \frac{Di}{Dt} = \frac{2}{0.05} = 40 \text{ A/s} \]

4. **Rearranging the formula to solve for \( L \)**:
\[ L = \frac{E}{\frac{Di}{Dt}} = \frac{8}{40} \]

5. **Calculate \( L \)**:
\[ L = \frac{8}{40} = 0.2 \text{ H} \]

6. **Conclusion**:
The coefficient of self-induction of the coil is \( L = 0.2 \text{ H} \).

### Final Answer:
The coefficient of self-induction of the coil is **0.2 Henry**.

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